A solid disk, spinning counter-clockwise, has a mass of $18 k g$ $18 kg$ and a radius of $\frac{4}{5} m$ $4/5 m$ . If a point on the edge of the disk is moving at $\frac{7}{3} \frac{m}{s}$ $7/3 m/s$ in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

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Answer 1 · 2017-08-10T07:42:57

The angular momentum is $= 16.8 k g m^{2} s^{- 1}$ $=16.8kgm^2s^-1$ and the angular velocity is $= 2.92 r a d s^{- 1}$ $=2.92rads^-1$

The angular velocity is

$omega=(Deltatheta)/(Deltat)$

$v=r*((Deltatheta)/(Deltat))=r omega$

$omega=v/r$

where,

$v=7/3ms^(-1)$

$r=4/5m$

So,

$omega=(7/3)/(4/5)=35/12=2.92rads^-1$

The angular momentum is $L=Iomega$

For a solid disc, $I=(mr^2)/2$

So, $I=18*(4/5)^2/2=144/25kgm^2$

The angular momentum is

$L=144/25*2.92=16.8kgm^2s^-1$

A solid disk, spinning counter-clockwise, has a mass of 18 kg18kg18 kg and a radius of 4/5 m45m4/5 m. If a point on the edge of the disk is moving at 7/3 m/s73ms7/3 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?